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I am certain there is already some algorithm that does what I need, but I am not sure what phrase to Google, or what is the algorithm category.
Here is my problem: I have a polyhedron made up by several contacting blocks (hyperslabs), i. e. the edges are axis aligned and the angles between edges are 90°. There may be holes inside the polyhedron.
I want to break up this concave polyhedron in as little convex rectangular axis-aligned whole blocks are possible (if the original polyhedron is convex and has no holes, then it is already such a block, and therefore, the solution). To illustrate, some 2-D images I made (but I need the solution for 3-D, and preferably, N-D):
I have this geometry:
One possible breakup into blocks is this:
But the one I want is this (with as few blocks as possible):
I have the impression that an exact algorithm may be too expensive (is this problem NP-hard?), so an approximate algorithm is suitable.
One detail that maybe make the problem easier, so that there could be a more appropriated/specialized algorithm for it is that all edges have sizes multiple of some fixed value (you may think all edges sizes are integer numbers, or that the geometry is made up by uniform tiny squares, or voxels).
Background: this is the structured grid discretization of a PDE domain.
What algorithm can solve this problem? What class of algorithms should I
search for?
Update: Before you upvote that answer, I want to point out that my answer is slightly off-topic. The original poster have a question about the decomposition of a polyhedron with faces that are axis-aligned. Given such kind of polyhedron, the question is to decompose it into convex parts. And the question is in 3D, possibly nD. My answer is about the decomposition of a general polyhedron. So when I give an answer with a given implementation, that answer applies to the special case of polyhedron axis-aligned, but it might be that there exists a better implementation for axis-aligned polyhedron. And when my answer says that a problem for generic polyhedron is NP-complete, it might be that there exists a polynomial solution for the special case of axis-aligned polyhedron. I do not know.
Now here is my (slightly off-topic) answer, below the horizontal rule...
The CGAL C++ library has an algorithm that, given a 2D polygon, can compute the optimal convex decomposition of that polygon. The method is mentioned in the part 2D Polygon Partitioning of the manual. The method is named CGAL::optimal_convex_partition_2. I quote the manual:
This function provides an implementation of Greene's dynamic programming algorithm for optimal partitioning [2]. This algorithm requires O(n4) time and O(n3) space in the worst case.
In the bibliography of that CGAL chapter, the article [2] is:
[2] Daniel H. Greene. The decomposition of polygons into convex parts. In Franco P. Preparata, editor, Computational Geometry, volume 1 of Adv. Comput. Res., pages 235–259. JAI Press, Greenwich, Conn., 1983.
It seems to be exactly what you are looking for.
Note that the same chapter of the CGAL manual also mention an approximation, hence not optimal, that run in O(n): CGAL::approx_convex_partition_2.
Edit, about the 3D case:
In 3D, CGAL has another chapter about Convex Decomposition of Polyhedra. The second paragraph of the chapter says "this problem is known to be NP-hard [1]". The reference [1] is:
[1] Bernard Chazelle. Convex partitions of polyhedra: a lower bound and worst-case optimal algorithm. SIAM J. Comput., 13:488–507, 1984.
CGAL has a method CGAL::convex_decomposition_3 that computes a non-optimal decomposition.
I have the feeling your problem is NP-hard. I suggest a first step might be to break the figure into sub-rectangles along all hyperplanes. So in your example there would be three hyperplanes (lines) and four resulting rectangles. Then the problem becomes one of recombining rectangles into larger rectangles to minimize the final number of rectangles. Maybe 0-1 integer programming?
I think dynamic programming might be your friend.
The first step I see is to divide the polyhedron into a trivial collection of blocks such that every possible face is available (i.e. slice and dice it into the smallest pieces possible). This should be trivial because everything is an axis aligned box, so k-tree like solutions should be sufficient.
This seems reasonable because I can look at its cost. The cost of doing this is that I "forget" the original configuration of hyperslabs, choosing to replace it with a new set of hyperslabs. The only way this could lead me astray is if the original configuration had something to offer for the solution. Given that you want an "optimal" solution for all configurations, we have to assume that the original structure isn't very helpful. I don't know if it can be proven that this original information is useless, but I'm going to make that assumption in this answer.
The problem has now been reduced to a graph problem similar to a constrained spanning forest problem. I think the most natural way to view the problem is to think of it as a graph coloring problem (as long as you can avoid confusing it with the more famous graph coloring problem of trying to color a map without two states of the same color sharing a border). I have a graph of nodes (small blocks), each of which I wish to assign a color (which will eventually be the "hyperslab" which covers that block). I have the constraint that I must assign colors in hyperslab shapes.
Now a key observation is that not all possibilities must be considered. Take the final colored graph we want to see. We can partition this graph in any way we please by breaking any hyperslab which crosses the partition into two pieces. However, not every partition is meaningful. The only partitions that make sense are axis aligned cuts, which always break a hyperslab into two hyperslabs (as opposed to any more complicated shape which could occur if the cut was not axis aligned).
Now this cut is the reverse of the problem we're really trying to solve. That cutting is actually the thing we did in the first step. While we want to find the optimal merging algorithm, undoing those cuts. However, this shows a key feature we will use in dynamic programming: the only features that matter for merging are on the exposed surface of a cut. Once we find the optimal way of forming the central region, it generally doesn't play a part in the algorithm.
So let's start by building a collection of hyperslab-spaces, which can define not just a plain hyperslab, but any configuration of hyperslabs such as those with holes. Each hyperslab-space records:
The number of leaf hyperslabs contained within it (this is the number we are eventually going to try to minimize)
The internal configuration of hyperslabs.
A map of the surface of the hyperslab-space, which can be used for merging.
We then define a "merge" rule to turn two or more adjacent hyperslab-spaces into one:
Hyperslab-spaces may only be combined into new hyperslab-spaces (so you need to combine enough pieces to create a new hyperslab, not some more exotic shape)
Merges are done simply by comparing the surfaces. If there are features with matching dimensionalities, they are merged (because it is trivial to show that, if the features match, it is always better to merge hyperslabs than not to)
Now this is enough to solve the problem with brute force. The solution will be NP-complete for certain. However, we can add an additional rule which will drop this cost dramatically: "One hyperslab-space is deemed 'better' than another if they cover the same space, and have exactly the same features on their surface. In this case, the one with fewer hyperslabs inside it is the better choice."
Now the idea here is that, early on in the algorithm, you will have to keep track of all sorts of combinations, just in case they are the most useful. However, as the merging algorithm makes things bigger and bigger, it will become less likely that internal details will be exposed on the surface of the hyperslab-space. Consider
+===+===+===+---+---+---+---+
| : : A | X : : : :
+---+---+---+---+---+---+---+
| : : B | Y : : : :
+---+---+---+---+---+---+---+
| : : | : : : :
+===+===+===+ +---+---+---+
Take a look at the left side box, which I have taken the liberty of marking in stronger lines. When it comes to merging boxes with the rest of the world, the AB:XY surface is all that matters. As such, there are only a handful of merge patterns which can occur at this surface
No merges possible
A:X allows merging, but B:Y does not
B:Y allows merging, but A:X does not
Both A:X and B:Y allow merging (two independent merges)
We can merge a larger square, AB:XY
There are many ways to cover the 3x3 square (at least a few dozen). However, we only need to remember the best way to achieve each of those merge processes. Thus once we reach this point in the dynamic programming, we can forget about all of the other combinations that can occur, and only focus on the best way to achieve each set of surface features.
In fact, this sets up the problem for an easy greedy algorithm which explores whichever merges provide the best promise for decreasing the number of hyperslabs, always remembering the best way to achieve a given set of surface features. When the algorithm is done merging, whatever that final hyperslab-space contains is the optimal layout.
I don't know if it is provable, but my gut instinct thinks that this will be an O(n^d) algorithm where d is the number of dimensions. I think the worst case solution for this would be a collection of hyperslabs which, when put together, forms one big hyperslab. In this case, I believe the algorithm will eventually work its way into the reverse of a k-tree algorithm. Again, no proof is given... it's just my gut instinct.
You can try a constrained delaunay triangulation. It gives very few triangles.
Are you able to determine the equations for each line?
If so, maybe you can get the intersection (points) between those lines. Then if you take one axis, and start to look for a value which has more than two points (sharing this value) then you should "draw" a line. (At the beginning of the sweep there will be zero points, then two (your first pair) and when you find more than two points, you will be able to determine which points are of the first polygon and which are of the second one.
Eg, if you have those lines:
verticals (red):
x = 0, x = 2, x = 5
horizontals (yellow):
y = 0, y = 2, y = 3, y = 5
and you start to sweep through of X axis, you will get p1 and p2, (and we know to which line-equation they belong ) then you will get p3,p4,p5 and p6 !! So here you can check which of those points share the same line of p1 and p2. In this case p4 and p5. So your first new polygon is p1,p2,p4,p5.
Now we save the 'new' pair of points (p3, p6) and continue with the sweep until the next points. Here we have p7,p8,p9 and p10, looking for the points which share the line of the previous points (p3 and p6) and we get p7 and p10. Those are the points of your second polygon.
When we repeat the exercise for the Y axis, we will get two points (p3,p7) and then just three (p1,p2,p8) ! On this case we should use the farest point (p8) in the same line of the new discovered point.
As we are using lines equations and points 2 or more dimensions, the procedure should be very similar
ps, sorry for my english :S
I hope this helps :)
I'm looking for an algorithm to find the best fit between a cloud of points and a sphere.
That is, I want to minimise
where C is the centre of the sphere, r its radius, and each P a point in my set of n points. The variables are obviously Cx, Cy, Cz, and r. In my case, I can obtain a known r beforehand, leaving only the components of C as variables.
I really don't want to have to use any kind of iterative minimisation (e.g. Newton's method, Levenberg-Marquardt, etc) - I'd prefer a set of linear equations or a solution explicitly using SVD.
There are no matrix equations forthcoming. Your choice of E is badly behaved; its partial derivatives are not even continuous, let alone linear. Even with a different objective, this optimization problem seems fundamentally non-convex; with one point P and a nonzero radius r, the set of optimal solutions is the sphere about P.
You should probably reask on an exchange with more optimization knowledge.
You might find the following reference interesting but I would warn you
that you will need to have some familiarity with geometric algebra -
particularly conformal geometric algebra to understand the
mathematics. However, the algorithm is straight forward to implement with
standard linear algebra techniques and is not iterative.
One caveat, the algorithm, at least as presented fits both center and
radius, you may be able to work out a way to constrain the fit so the radius is constrained.
Total Least Squares Fitting of k-Spheres in n-D Euclidean Space Using
an (n+ 2)-D Isometric Representation. L Dorst, Journal of Mathematical Imaging and Vision, 2014 p1-21
Your can pull in a copy from
Leo Dorst's researchgate page
One last thing, I have no connection to the author.
Short description of making matrix equation could be found here.
I've seen that WildMagic Library uses iterative method (at least in version 4)
You may be interested by the best fit d-dimensional sphere, i.e. minimizing the variance of the population of the squared distances to the center; it has a simple analytical solution (matrix calculus): see the appendix of the open access paper of Cerisier et al. in J. Comput. Biol. 24(11), 1134-1137 (2017), https://doi.org/10.1089/cmb.2017.0061
It works when the data points are weighted (it works even for continuous distributions; as a by-product, when d=1, a well-known inequality is retrieved: the kurtosis is always greater than the squared skewness plus 1).
Difficult to do this without iteration.
I would proceed as follows:
find the overall midpoint, by averaging (X,Y,Z) coords for all points
with that result, find the average distance Ravg to the midpoint, decide ok or proceed..
remove points from your set with a distance too far from Ravg found in step 2
go back to step 1 (average points again, yielding a better midpoint)
Of course, this will require some conditions for (2) and (4) that depends on the quality of your points cloud !
Ian Coope has an interesting algorithm in which he linearized the problem using a change of variable. The fit is quite robust, and although it slightly redefines the condition of optimality, I've found it to be generally visually better, especially for noisy data.
A preprint of Coope's paper is available here: https://ir.canterbury.ac.nz/bitstream/handle/10092/11104/coope_report_no69_1992.pdf.
I found the algorithm to be very useful, so I implemented it in scikit-guess as skg.nsphere_fit. Let's say you have an (m, n) array p, consisting of M points of dimension N (here N=3):
r, c = skg.nsphere_fit(p)
The radius, r, is a scalar and c is be an n-vector containing the center.
There are two classes, let's call them X and O. A number of elements belonging to these classes are spread out in the xy-plane. Here is an example where the two classes are not linearly separable. It is not possible to draw a straight line that perfectly divides the Xs and the Os on each side of the line.
How to determine, in general, whether the two classes are linearly separable?. I am interested in an algorithm where no assumptions are made regarding the number of elements or their distribution. An algorithm of the lowest computational complexity is of course preferred.
If you found the convex hull for both the X points and the O points separately (i.e. you have two separate convex hulls at this stage) you would then just need to check whether any segments of the hulls intersected or whether either hull was enclosed by the other.
If the two hulls were found to be totally disjoint the two data-sets would be geometrically separable.
Since the hulls are convex by definition, any separator would be a straight line.
There are efficient algorithms that can be used both to find the convex hull (the qhull algorithm is based on an O(nlog(n)) quickhull approach I think), and to perform line-line intersection tests for a set of segments (sweepline at O(nlog(n))), so overall it seems that an efficient O(nlog(n)) algorithm should be possible.
This type of approach should also generalise to general k-way separation tests (where you have k groups of objects) by forming the convex hull and performing the intersection tests for each group.
It should also work in higher dimensions, although the intersection tests would start to become more challenging...
Hope this helps.
Computationally the most effective way to decide whether two sets of points are linearly separable is by applying linear programming. GLTK is perfect for that purpose and pretty much every highlevel language offers an interface for it - R, Python, Octave, Julia, etc.
Let's say you have a set of points A and B:
Then you have to minimize the 0 for the following conditions:
(The A below is a matrix, not the set of points from above)
"Minimizing 0" effectively means that you don't need to actually optimize an objective function because this is not necessary to find out if the sets are linearly separable.
In the end
() is defining the separating plane.
In case you are interested in a working example in R or the math details, then check this out.
Here is a naïve algorithm that I'm quite sure will work (and, if so, shows that the problem is not NP-complete, as another post claims), but I wouldn't be surprised if it can be done more efficiently: If a separating line exists, it will be possible to move and rotate it until it hits two of the X'es or one X and one O. Therefore, we can simply look at all the possible lines that intersect two X'es or one X and one O, and see if any of them are dividing lines. So, for each of the O(n^2) pairs, iterate over all the n-2 other elements to see if all the X'es are on one side and all the O's on the other. Total time complexity: O(n^3).
Linear perceptron is guaranteed to find such separation if one exists.
See: http://en.wikipedia.org/wiki/Perceptron .
You can probably apply linear programming to this problem. I'm not sure of its computational complexity in formal terms, but the technique is successfully applied to very large problems covering a wide range of domains.
Computing a linear SVM then determining which side of the computed plane with optimal marginals each point lies on will tell you if the points are linearly separable.
This is overkill, but if you need a quick one off solution, there are many existing SVM libraries that will do this for you.
As mentioned by ElKamina, Linear Perceptron is guaranteed to find a solution if one exists. This approach is not efficient for large dimensions. Computationally the most effective way to decide whether two sets of points are linearly separable is by applying linear programming.
A code with an example to solve using Perceptron in Matlab is here
In general that problem is NP-hard but there are good approximate solutions like K-means clustering.
Well, both Perceptron and SVM (Support Vector Machines) can tell if two data sets are separable linearly, but SVM can find the Optimal Hiperplane of separability. Besides, it can work with n-dimensional vectors, not only points.
It is used in applications such as face recognition. I recomend to go deep into this topic.
I'm writing an app that looks up points in two-dimensional space using a k-d tree. It would be nice, during development, to be able to "see" the nearest-neighbor zones surrounding each point.
In the attached image, the red points are points in the k-d tree, and the blue lines surrounding each point bound the zone where a nearest neighbor search will return the contained point.
The image was created thusly:
for each point in the space:
da = distance to nearest neighbor
db = distance to second-nearest neighbor
if absolute_value(da - db) < 4:
draw blue pixel
This algorithm has two problems:
(more important) It's slow on my (reasonably fast Core i7) computer.
(less important) It's sloppy, as you can see by the varying widths of the blue lines.
What is this "visualization" of a set of points called?
What are some good algorithms to create such a visualization?
This is called a Voronoi Diagram and there are many excellent algorithms for generating them efficiently. The one I've heard about most is Fortune's algorithm, which runs in time O(n log n), though others algorithms exist for this problem.
Hope this helps!
Jacob,
hey, you found an interesting way of generating this Voronoi diagram, even though it is not so efficient.
The less important issue first: the varying thickness boundaries that you get, those butterfly shapes, are in fact the area between the two branches of an hyperbola. Precisely the hyperbola given by the equation |da - db| = 4. To get a thick line instead, you have to modify this criterion and replace it by the distance to the bisector of the two nearest neighbors, let A and B; using vector calculus, | PA.AB/||AB|| - ||AB||/2 | < 4.
The more important issue: there are two well known efficient solutions to the construction of the Voronoi diagram of a set of points: Fortune's sweep algorithm (as mentioned by templatetypedef) and Preparata & Shamos' Divide & Conquer solutions. Both run in optimal time O(N.Lg(N)) for N points, but aren't so easy to implement.
These algorithm will construct the Voronoi diagram as a set of line segments and half-lines. Check http://en.wikipedia.org/wiki/Voronoi_diagram.
This paper "Primitives for the manipulation of general subdivisions and the computation of Voronoi" describes both algorithms using a somewhat high-level framework, caring about all implementation details; the article is difficult but the algorithms are implementable.
You may also have a look at "A straightforward iterative algorithm for the planar Voronoi diagram", which I never tried.
A totally different approach is to directly build the distance map from the given points for example by means of Dijkstra's algorithm: starting from the given points, you grow the boundary of the area within a given distance from every point and you stop growing when two boundaries meet. [More explanations required.] See http://1.bp.blogspot.com/-O6rXggLa9fE/TnAwz4f9hXI/AAAAAAAAAPk/0vrqEKRPVIw/s1600/distmap-20-seed4-fin.jpg
Another good starting point (for efficiently computing the distance map) can be "A general algorithm for computing distance transforms in linear time".
From personal experience: Fortune's algorithm is a pain to implement. The divide and conquer algorithm presented by Guibas and Stolfi isn't too bad; they give detailed pseudocode that's easy to transcribe into a procedural programming language. Both will blow up if you have nearly degenerate inputs and use floating point, but since the primitives are quadratic, if you can represent coordinates as 32-bit integers, then you can use 64 bits to carry out the determinant computations.
Once you get it working, you might consider replacing your kd-tree algorithms, which have a Theta(√n) worst case, with algorithms that work on planar subdivisions.
You can find a great implementation for it at D3.js library: http://mbostock.github.com/d3/ex/voronoi.html
The problem of finding the intersection/union of 'N' discs/circles on a flat plane was first proposed by M. I. Shamos in his 1978 thesis:
Shamos, M. I. “Computational Geometry” Ph.D. thesis, Yale Univ., New Haven, CT 1978.
Since then, in 1985, Micha Sharir presented an O(n log2n) time and O(n) space deterministic algorithm for the disc intersection/union problem (based on modified Voronoi diagrams): Sharir, M. Intersection and closest-pair problems for a set of planar discs. SIAM .J Comput. 14 (1985), pp. 448-468.
In 1988, Franz Aurenhammer presented a more efficient O(n log n) time and O(n) space algorithm for circle intersection/union using power diagrams (generalizations of Voronoi diagrams): Aurenhammer, F. Improved algorithms for discs and balls using power diagrams. Journal of Algorithms 9 (1985), pp. 151-161.
Earlier in 1983, Paul G. Spirakis also presented an O(n^2) time deterministic algorithm, and an O(n) probabilistic algorithm: Spirakis, P.G. Very Fast Algorithms for the Area of the Union of Many Circles. Rep. 98, Dept. Comput. Sci., Courant Institute, New York University, 1983.
I've been searching for any implementations of the algorithms above, focusing on computational geometry packages, and I haven't found anything yet. As neither appear trivial to put into practice, it would be really neat if someone could point me in the right direction!
Perhaps a constructive solid geometry (CSG) package would have a surface-area feature for overlapping circle primitives?
I've spent some time looking at the algorithms for computing the union of a set of balls - as you mention, it is done by using generalized Voronoi diagrams.
The CGAL library has a package that implements a union of balls. This is one dimension higher than what you're interested in, and it doesn't handle intersections. so probably no cigar.
If you're working in 2 dimensions, the problem is equivalent to finding a convex hull in 3 dimensions. You could use the convex hull stuff from CGAL or another package.
If you're looking for implementations of the specific algorithms you've mentioned, I can't help you. But if you just want to find the power diagram that allows you to easily compute unions and intersections, it might be easier than you think to roll your own using a 3D convex hull algorithm.
Sorry for the half baked answer, but we may as well start somewhere and see how much flexibility you have.
Edit: There's also a CGAL package for 2D power diagrams: http://www.cgal.org/Manual/last/doc_html/cgal_manual/Apollonius_graph_2/Chapter_main.html.
Also, #RGrey has found a CGAL library for calculating 2D booleans for generalized polygons, including circles at http://www.cgal.org/Manual/3.5/doc_html/cgal_manual/Boolean_set_operations_2/Chapter_main.html.
There is no 2D power diagram implementation in CGAL. The link suggested above (http://www.cgal.org/Manual/last/doc_html/cgal_manual/Apollonius_graph_2/Chapter_main.html) is for the additive weighted Voronoi diagram, which uses the euclid + weight, instead of euclid^2 + weight (as Power Diagrams do).